- 7.1 | Area of a Region Between Two Curves
- 7.3 | Volume: The Shell Method
- 7.4 | Arc Length and Surfaces of Revolution
- 7.5 | Work

**Volume of a Solid of Revolution**(Region revolved about the \(x\)-axis)

Let \(f\) be a continuous nonnegative function on \([a,b],\) and let \(R\) be the region under the graph of \(f\) on the interval \([a,b].\) The volume of the solid of revolution generated by revolving \(R\) about the \(x\)-axis is

$$

V = \lim_{n \rightarrow \infty} \sum_{k =1}^n \pi [f(c_k)]^2 \; \Delta x = \int_a^b \pi[f(x)]^2 \; dx

$$

$$

\int_a^b \pi [f(x)]^2-\pi[g(x)]^2 \; dx

$$

where \(f(x) \geq g(x) >0\) on \([a,b]\). \(f(x)\) represents the outer radius and \(g(x)\) represents the inner radius on \([a,b]\). When \(g(x)=0\) this formula is referred to as the

**disk method**, otherwise this is referred to as the

**washer method.**

Set up the integral(s) that represent the volume of the solid of revolution generated by revolving the enclosed region on the left about the \(x\)-axis.

$$

\begin{aligned}

\int_0^1 \pi(2)^2 \; dx + \int_1^2 \pi(x+1)^2-\pi \left( \sqrt{1-(x-2)^2} \right)^2\; dx \\

+ \int_2^3 \pi (3)^2-\pi \left( \sqrt{1-(x-2)^2} \right)^2 \; dx

\end{aligned}

$$

where

- on the interval \([0,1]\), the outer radius is given by \(f(x) = 2\) and the inner radius is given by \(g(x) = 0\)
- on the interval \([1,2]\), the outer radius is given by \(f(x) = x+1\) and the inner radius is given by \(g(x) = \sqrt{1-(x-2)^2}\).
- on the interval \([2,3]\), the outer radius is given by \(f(x) = 3\) and the inner radius is given by \(g(x) = \sqrt{1-(x-2)^2}\).

A ball of radius 17 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid.

Notice that this is equivalent to the standard solids of revolution problem:

Find the volume of the solid of revolution obtained by revolving the region bounded by the graphs of

$$

x = \sqrt{17^2-y^2}, \text{ and } x = 5

$$

about the \(y\)-axis.

The volume of the solid of revolution, by the washer method, is

$$

\begin{aligned}

V & = \int_{-\sqrt{264}}^{\sqrt{264}} \pi\left(\sqrt{289-y^2}\right)^2-25 \pi \; dy \\

& =\int_{-\sqrt{264}}^{\sqrt{264}}\pi (264-y^2) \; dy \\

& = \pi \left(264y-\frac{y^3}{3}\right)\Big|_{-\sqrt{264}}^{\sqrt{264}} \\

& =704 \sqrt{66} \pi \approx 17967.8

\end{aligned}

$$

**Volume of a Solid with Known Cross Section**Let \(S\) be a solid bounded by planes that are perpendicular to the \(x\)-axis at \(x=a\) and \(x=b\). If the cross-sectional area of \(S\) at any point \(x\) in \([a,b]\) is \(A(x)\), where \(A\) is continuous on \([a,b]\), then the volume of \(S\) is

$$

V = \lim_{n \rightarrow \infty} \sum_{k=1}^n A(c_k) \Delta x = \int_a^b A(x) \; dx

$$

Imagine the base of the solid centered on the \(xy\)-plane. Then the equation of the circle along the base is given by \(x^2+y^2 = 4\). Furthermore, the length of the base of the cross sectional region can now be seen to be \(4y\). Therefore, the area of the cross sectional region (square) is \(16y^2\). Since this cross sectional region is being moved along the \(x\)-axis, we know that the integral representing the volume of the solid must be expressed entirely in the variable \(x\) and since \(y = \pm\sqrt{4-x^2}\), we have

$$

V = \int_{-2}^2 16(4-x^2) \; dx = \frac{512}{3}

$$

The cross-sections of the solid, taken perpendicular to the \(y\)-axis, are square. In which case, the volume that these cross-sections contribute to the overall volume of the solid is given by

$$

\begin{aligned}

\Delta V & = l^2 \cdot h \\

& = (\sqrt{y} – (-\sqrt{y}))^2 \cdot \Delta y \\

& = (2 \sqrt{y})^2 \; \Delta y \\

& = 4 |y| \; \Delta y \\

& = 4 y \; \Delta y

\end{aligned}

$$

for \(0 \leq y \leq 1\).

So the volume of the solid is given by

$$

V = \int_0^1 4y \; dy = 2y^2 \Big|_0^1 = 2

$$